$A$ body crosses the topmost point of a vertical circle with critical speed. Its centripetal acceleration,when the string is horizontal,will be $....g$

$A$ particle starts sliding from the top of a sphere of diameter $42 \, m$. At what height from the bottom of the sphere will the particle lose contact with the sphere?

The minimum horizontal speed with which a body must be projected, so that it goes around a smooth vertical circular track of radius $4 \,m$ is $(g=9.8 \,ms^{-2})$. (in $\,ms^{-1}$)

$A$ ball whose size is slightly smaller than the width of a tube of radius $R = 2.5 \, m$ is projected from the bottommost point of a smooth tube fixed in a vertical plane with a velocity of $v_0 = 10 \, m/s$. Let $N_1$ be the normal reaction exerted by the inner side of the tube and $N_2$ be the normal reaction exerted by the outer side of the tube on the ball. Which of the following statements is correct?

$A$ stone of mass $m$ is tied to a string and is moved in a vertical circle of radius $r$ making $n$ revolutions per minute. The total tension in the string when the stone is at its lowest point is

$A$ particle with charge $Q$ coulomb, tied at the end of an inextensible string of length $R$ metre, revolves in a vertical plane. At the centre of the circular trajectory, there is a fixed charge of magnitude $Q$ coulomb. The mass of the moving charge $M$ is such that $Mg = \frac{Q^2}{4 \pi \epsilon_0 R^2}$. If at the highest position of the particle, the tension of the string just vanishes, the horizontal velocity at the lowest point has to be